Dynamic Systems - DE with Unit Impulse

In summary: You can't just remove things that you don't know (or don't want to see) and expect it all to remain the same.In summary, the given differential equations can be solved for y(t) by evaluating the integral and simplifying using c >> |a|. The simplified solution for y(t) is bδ(t), and the resulting derivative satisfies the original DE approximately. However, removing information from the solution will result in a different form that may not fully satisfy the DE.
  • #1
Graphite
7
0

Homework Statement


[tex]y(t) = e^{a(t-t_{0})} y(t_{0}) + \int_{t_{0}}^{t} e^{a(t-\tau)}b \dot{u} (\tau) d\tau[/tex]
[tex]u(t) = \delta(t) = \frac{1}{2} c^{3} t^{2} e^{-ct}[/tex]
where c >> |a|, t_0 = 0, and y(0) = 0
Find y(t) and represent the unit impulse, delta, in the solution. The remaining
terms should not contain c.

Homework Equations

The Attempt at a Solution


Evaluating the integral, I got
[tex]y = \frac{1}{2}bc^{3} \left( e^{-ct} \left(\frac{(ct^{2}-2t)(a+c)^{2}+(2ct-2)(a+c)+2c}{(a+c)^{3}} \right) + e^{at} \frac{2a}{(a+c)^{3}} \right) [/tex]
However, I'm stuck at trying to substitute in delta to get a function that does not have the variable c in it. I'm assuming that since c >> |a|, then the terms with (a+c) will simplify to just c
[tex]y = \frac{1}{2}b \left( e^{-ct} \left((ct^{2}-2t)(c)^{2}+(2ct-2)(c)+2c \right) + e^{at} {2a}\right) [/tex]
When I simplify it further, I am still unable to get it in a form without c. I'm thinking that I made a mistake somewhere in calculating the integral, but Matlab is giving me the similar result so I"m not sure where the problem is.
 
Physics news on Phys.org
  • #2
I get something different when I evaluate the integral, but nevermind that. Try simplifying using c >> |a| before you evaluate the integral.
 
  • #3
Doing that, I'm getting
[tex]y = \frac{1}{2} b c^{3} t^{2} e^{(a-c)t} = be^{at}\delta(t)[/tex]
which is much simpler.

I just realized that I forgot to post the DE at the beginning, which is
[tex]\dot{y} = ay + b \dot{u}[/tex]
When I try taking the derivative of y, I'm getting
[tex]\dot{y} = abe^{at}\delta(t) + be^{at}\dot{\delta}(t) = ay + be^{at} \dot{u}(t)[/tex]
I'm assuming that since [itex]\dot{u}(t)[/itex] has the term [itex]e^{-ct}[/itex], then the [itex]e^{at}[/itex] will disappear, which would satisfy the DE.
Why is it that if I were to simplify y at the beginning to
[tex]y = \frac{1}{2} b c^{3} t^{2} e^{(-c)t} = b\delta(t)[/tex]
and then taking the derivative,
[tex]\dot{y} = b\dot{u}(t)[/tex]
Why is it that now it does not satisfy the DE, since [itex]ay[/itex] is now missing? I'm curious as to why removing the variable a when it's being added/subtracted from c results in different forms depending on the step at which the simplification is done. Are all of the forms roughly equivalent to each other or something?

Thank you for the help.
 
  • #4
Graphite said:
Why is it that now it does not satisfy the DE, since [itex]ay[/itex] is now missing?
It does satisfy the DE, approximately, assuming ##c \gg |a|##. You can try plotting:
$$
\dot{y} = \frac{d}{dt}\left[be^{at} \delta(t)\right]\\
\dot{y} = \frac{d}{dt}\left[b \delta(t)\right]
$$
for some values of ##a,b,c##. Their graphs should be almost identical.

You have something "missing", because you've removed more information from the solution, comparatively - that has to be reflected somewhere.
 

FAQ: Dynamic Systems - DE with Unit Impulse

What is a dynamic system?

A dynamic system is a system that changes over time, where the current state of the system is influenced by its previous state and external inputs. These systems are often described using differential equations.

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. In the context of dynamic systems, it represents how the state of the system changes over time.

What is a unit impulse?

A unit impulse is a mathematical function that represents an instantaneous, infinitesimally short burst of energy. In the context of dynamic systems, it is often used as a forcing function to study the response of the system.

How is a unit impulse used in dynamic systems?

A unit impulse is used as a forcing function in dynamic systems to study the response of the system. By applying a unit impulse at different points in time, we can observe how the system reacts and how its state changes over time.

What are some real-world applications of dynamic systems with unit impulse?

Dynamic systems with unit impulse have many real-world applications, such as in electrical circuits, control systems, and signal processing. They are also used to model the behavior of physical systems, such as a bouncing ball or a swinging pendulum.

Similar threads

Replies
2
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
8
Views
2K
Replies
4
Views
3K
Replies
6
Views
2K
Replies
8
Views
2K
Back
Top